2 Answers
2

Nice answer BS. I was about to post something similar but I didn't have a proof of the non-linearity of the action of $k$-jets of diffeomorphisms. One additional remark that might help the OP is that you can put a vector space structure on $J^k_0({\mathbb R}^n, M)_x$ if you have additional structure on $M$. Basically you need to be able to determine a family of co-ordinates (actually $k$-jets of co-ordinates) that are related by linear transformations to avoid the non-linear action of diffeomorphisms when you change co-ordinates. Also you want to choose different co-ordinates at each point of $M$. Sufficient would be to choose at each $x \in M$ the $k$-jet of a diffeomorphism from $M$ to $T_xM$ sending $x$ to $0 \in T_xM$. For example if $M$ is Riemannian the $k$-jet of the inverse of the exponential map would do or if $M$ is a submanifold orthogonal projection onto the tangent subspace would work. In such a case composition with the chosen $k$-jet of a diffeomorphism defines a bijection
$$
J^k_0({\mathbb R}^n, M)_x \to J^k_0({\mathbb R}^n, T_xM)_0
$$
and the latter space is a vector space because $T_xM$ is a vector space.

Thank you for your answer.I have thought that the space $(T_k^rM)_x$ has vector space structure relative to tangent space $T_xM$. I will study about this more, considering your ansmer.
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user26296Sep 8 '12 at 18:28

I assume you ask for a "natural" vector space structure, one for which the natural group action of $r$-jets of diffeomorphisms of $M$ fixing $x$ is linear, and that the vector structure is smooth in standard coordinates.

Then the answer is negative for all $r>1$ (for $r=1$ it is of course positive, the space is $(T_x M)^k$).

For the proof, let $\phi:(\mathbb{R}^n,0)\to(\mathbb{R}^n,0)$ be a germ of diffeomorphism of the form $\phi(x)=x+h(x)$ with $h\neq 0$ homogeneous of degree $r$, and $f:(\mathbb{R}^k,0)\to(\mathbb{R}^n,0)$.

Then for $k\lt r$, $$D^k(\phi\circ f)(0)=D^kf(0),$$ and $$D^r(\phi\circ f)(0)=D^rf(0)+D^rh(Df(0),...,Df(0))$$.

Hence the action of $J_0^r\phi$ on $J_0^r(\mathbb{R}^k,\mathbb{R}^n)_0$ is tangent to the identity at the jet of the zero map $\mathbf{0}$, but is not the identity.

Now assume for contradiction that there is a "natural" vector space structure. Since the $\mathbf{0}$ is the only jet fixed by all $r$-jets of diffeomorphisms fixing $0$, it is the only possible zero of a natural vector space structure.

Now, by assumption the action of $J_0^r\phi$ is linearizable by a smooth ($C^1$ is enough) diffeomorphism $$\psi: (J_0^r(\mathbb{R}^k,\mathbb{R}^n)_0,\mathbf{0}) \to (\mathbb{R}^N,0),$$ but the only possible candidate for the linear map $\psi\circ J_0^r\phi\circ\psi^{-1}$ from $(\mathbb{R}^N,0)$ to itself is its derivative at 0, i.e. the identity, a contradiction.

PS: the natural structure on jets is a bit complicated. There is a sequence of fiber bundles $$T_k^r M\to T^{r-1}_k M\to...\to T^1_k M\to M,$$
where the rightmost map is a vector bundle, but the others are affine bundles under the vector bundles $P_k^r M \simeq S^r(\mathbb{R}^k)^*\otimes TM$ of jets in $T_k^r M$ which map to constant jets in $T^{r-1}_k M$.

To see that the $P_k^r M$ are vector bundles, one may observe that a $r$-jet of diffeomorphism $J_x^r\phi$ of $M$ acts on $(P_k^r M)_x$ only through $J_x^1\phi$.

It is remarkable that the choice of an affine connection on $M$, which may be viewed as a splitting of $$0\to P_1^2 M\to T_1^2 M\to T^1_1M=TM\to 0,$$
is sufficient to put a vector bundle structure on all jet bundles, by using covariant derivatives, see this wikipedia article and its references for more details, notably on the important Cartan distribution on jet bundles.